Question

A celebrated Mark Twain story has motivated contestants in the Calaveras County Jumping Frog Jubilee, where frog jumps as long as 2.20 m have been recorded. If a frog jumps 2.20 m and the launch angle is 36.5°, find the frog's launch speed and the time the frog spends in the air. Ignore air resistance.
(a)the frog's launch speed (in m/s)

(b)the time the frog spends in the air (in s)

Answer 1

The frog's launch speed and the time spends in the air are 22.5m/s and 2.73s respectively.

To find the answer, we need to know about the time of flight and range of projectile motion.

What's the expression of range of a projectile motion?

• Range = U²× sin(2θ)/g
• U= initial velocity, θ= angle of projectile and g= acceleration due to gravity
• U=√{Range×g/sin(2θ)}
• Here, range= 2.20m, = 36.5°
• U= √{2.20×9.8/sin(73)}

U= √{2.20×9.8/sin(73)} = 22.5m/s

What's the expression of time of flight in projectile motion?

• Time of flight= (2×U×sinθ)/g
• So, T= (2×22.5×sin36.5°)/9.8

= 2.73 s

Thus, we can conclude that the frog's launch speed and the time spends in the air are 22.5m/s and 2.73s respectively.

Learn more about the range and time period of projectile motion here:

brainly.com/question/24136952

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